Theorem sbc3or 44552

Description: sbcor 3839 with a 3-disjuncts. This proof is sbc3orgVD 44871 automatically translated and minimized. (Contributed by Alan Sare, 31-Dec-2011.) (Revised by NM, 24-Aug-2018.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
sbc3or	⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒))

Proof of Theorem sbc3or

Step	Hyp	Ref	Expression
1		sbcor 3839	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ∨ [𝐴 / 𝑥]𝜒))
2		df-3or 1088	. . . . 5 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
3	2	bicomi 224	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ (𝜑 ∨ 𝜓 ∨ 𝜒))
4	3	sbcbii 3846	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ [𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒))
5		sbcor 3839	. . . 4 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓))
6	5	orbi1i 914	. . 3 ⊢ (([𝐴 / 𝑥](𝜑 ∨ 𝜓) ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
7	1, 4, 6	3bitr3i 301	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
8		df-3or 1088	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
9	7, 8	bitr4i 278	1 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒))

Syntax hints:   ↔ wb 206   ∨ wo 848   ∨ w3o 1086  [wsbc 3788

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3or 1088  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-v 3482  df-sbc 3789

This theorem is referenced by:  sbcoreleleq  44555